Question

If two curves $x^2 - 4y^2 = 2$ and $8x^2 = 40 - my^2$ are orthogonal to each other then m =

• A. 2
• B. 16
• C. $\frac{1}{\sqrt{2}}$
• D. 4

Hint:

Orthogonal curves have perpendicular tangents at every point of intersection.

Answer 1

The Correct Option is B

Step 1: Concept
Two curves are orthogonal if the product of their slopes at the intersection point is $-1$ ($m_1 m_2 = -1$).

Step 2: Meaning
1) $2x - 8y y' = 0 \implies m_1 = x/4y$.
2) $16x = -2m y y' \implies m_2 = -8x/my$.

Step 3: Analysis
$m_1 m_2 = (x/4y)(-8x/my) = -2x^2 / my^2 = -1 \implies 2x^2 = my^2$.
From curves: $x^2 = 4y^2 + 2$. Substitute into condition: $2(4y^2 + 2) = my^2 \implies 8y^2 + 4 = my^2$.

Step 4: Conclusion
Comparing with the second curve equation $8x^2 + my^2 = 40$, we solve for the constant ratio condition of orthogonal conic sections, yielding $m = 16$. Final Answer: (B)